Point $D$ is on side $AC$ of triangle $ABC$, $\angle ABD=15^{\circ}$ and  $\angle DBC=50^{\circ}$. What is the measure of angle $BAD$, in degrees?

[asy]draw((-43,0)--(43,0)--(43,40)--cycle);
draw((-4,0)--(43,40));
draw((39,4)--(39,0));
draw((39,4)--(43,4));
draw((-1,36)--(22.5,26),Arrow);
label("$15^{\circ}$",(-1,36),W);
label("$50^{\circ}$",(36,30),S);
label("D",(-4,0),S);
label("A",(-43,0),W);
label("B",(43,40),N);
label("C",(43,0),E);
[/asy]
Solution: From triangle $ABC$, we have $\angle A = 180^\circ - \angle C - \angle ABC = 180^\circ - 90^\circ - (50^\circ + 15^\circ) = 90^\circ - 65^\circ = \boxed{25^\circ}$.